# 实现 pow(x, n) ，即计算 x 的 n 次幂函数。
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#  示例 1:
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#  输入: 2.00000, 10
# 输出: 1024.00000
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#  示例 2:
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#  输入: 2.10000, 3
# 输出: 9.26100
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#  示例 3:
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#  输入: 2.00000, -2
# 输出: 0.25000
# 解释: 2-2 = 1/22 = 1/4 = 0.25
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#  说明:
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#  -100.0 < x < 100.0
#  n 是 32 位有符号整数，其数值范围是 [−231, 231 − 1] 。
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#  Related Topics 数学 二分查找


# leetcode submit region begin(Prohibit modification and deletion)
class Solution:
    def myPow(self, x: float, n: int) -> float:
        # cheat
        # return x ** n
        if n < 0:
            return 1 / self.helper(x, -n)
        else:
            return self.helper(x, n)

    def helper(self, x, n):
        # recursion terminator
        if n == 0: return 1
        # process data
        # conquer sub problems
        half = self.helper(x, n // 2)
        # process final result
        if n % 2 == 0:
            return half * half
        else:
            return half * half * x

# leetcode submit region end(Prohibit modification and deletion)
